Using Frobenius Reciprocity, derive the character table of S_n from that of A_n when

(a) n = 3

(b) n = 4

(c) n = 5

Printable View

- Dec 15th 2009, 03:14 PMakc2010characters of S_n from A_n
Using Frobenius Reciprocity, derive the character table of S_n from that of A_n when

(a) n = 3

(b) n = 4

(c) n = 5 - Dec 15th 2009, 10:10 PMaliceinwonderland
This is for n=5.

The character table for S_5 is

$\displaystyle \begin{matrix}

\text{size}&1 & 10 & 20 & 30 &24 & 15 &20\\

&1 & (12) & (123) & (1234) & (12345) &(12)(34) & (12)(345)\\

U & 1 & 1 & 1 & 1 & 1 &1 &1\\

U' & 1 & -1 & 1 & -1 & 1 & 1 & -1\\

V & 4 & 2 & 1 & 0 & -1 & 0 &-1\\

V' & 4 & -2 & 1 & 0 & -1 & 0 & 1 \\

\wedge^2V & 6 & 0 & 0 & 0 & 1 & -2 &0 \\

W & 5 & 1 & -1 & -1 & 0 & 1&1\\

W' & 5 & -1 & -1 & 1 &0 &1 & -1 \\

\end{matrix}$

The character table for A_5 is here.

I followed the notations of the book, Fulton's "Representation Theory : A first course".

Note that the conjugacy class (12345) in S_5 with size 24 splits into two conjugacy classes (12345) and (21345) in A_5 with size 12, respectively. Note also that the conjugacy classes (12), (1234) and (12)(345) are not available in A_5.

Let H=A_5 and G=S_5. Then, by using Frobenius's reciprocity

Res U = U, Res U' = U, Ind U = U $\displaystyle \oplus$ U',

Res V = V, Res V' = V, Ind V = V $\displaystyle \oplus$ V',

Res W = W, Res W' = W, Ind W = W $\displaystyle \oplus$ W', and

Res $\displaystyle \wedge^2V$ = Y $\displaystyle \oplus$ Z, Ind Y = Ind Z = $\displaystyle \wedge^2V$.

The last line is obtained by using the lemma,

$\displaystyle \chi^G(g) = \sum_{i=1}^s|G:H|\frac{k_i}{l}\chi(h_i)$,

where $\displaystyle l$ is the order of the conjugacy class in G of g,

$\displaystyle h_1, h_2, \cdots, h_s$ are representatives of s conjugacy classes of H,

$\displaystyle k_1, k_2, \cdots , k_s$ be the orders of these classes.